我有數學的問題~急x)
1.(a) Prove that the sequence defined by x1= 3 and Xn+1 =1/4−xn
converges.
(b) Now that we know limxn exists, explain why limxn+1 must also exist
and equal the same value.
(c) Take the limit of each side of the recursive equation in part (a) of this
exercise to explicitly compute limxn.
2.Let x1= 2, and definexn+1 =1/2(xn+2/xn)
.
(a) Show that x上2下n is always greater than 2, and then use this to prove thatxn−xn+1 ≥ 0. Conclude that limXn=√2.
(b) Modify the sequence (xn) so that it converges to √c.
Update:is 1/(4- xn)
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Verified answer
(1/4)-(xn) or 1/(4- xn) ?
2013-10-06 22:50:39 補充：
1. (a)
(1)Show that xn in [1,3] by M.I.
n=1, obviously
Assume that xk in [1,3], then 4-xk in [1,3], x(k+1)=1/(4-xk) in [1/3, 1]
so, x(k+1) in [1,3]
(2)Show that xn is decreasing.
x(n+1)-x(n)= 1/(4-xn) - xn = (xn^2-4xn+1)/(4-xn)
=[(xn-2)^2 -3]/(4-xn) < 0 ( since xn in [1,3] )
From (1),(2) then the sequence {xn} is decreasing and bounded,
so that {xn} is convergents.
1. (b)
Set lim(n-> ∞) xn =L, then for all ε > 0, there exists N>0, such that
when n > N, then |x(n)-L| < ε , so |x(n+1) - L| < ε, ie. lim(n-> ∞) x(n+1)= L
1. (c)
lim(n-> ∞) x(n+1) = lim(n-> ∞) 1/(4-xn), then L= 1/(4-L)
L^2- 4L +1=0, L=2-√3 (2+√3 is rejected since xn in [1,3] )
2. x1=2, x(n+1)= (xn+ 2/xn)/2, then xn > 0
(a) x(n+1)= (xn+ 2/xn)/2 > = √2 (since AP > = GP), so (xn)^2 > = 2
x(n+1)-xn= (2/xn - xn)/2 = (2-xn^2)/(2xn) < = 0
so {xn} is bounded and decreasing, then {xn} is convergent.
Let {xn} converge to L, then
lim x(n+1)= lim (xn+ 2/xn)/2 ( n -> inf is omitted)
L= (L + 2/L)/2 , so L=√2
(b) x1=c+1, x(n+1)= (xn + c/xn)/2, c > 0, then lim (xn)= √c
看不懂啦!
不會寫國語嗎?